- #1

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## Homework Statement

what is the limit of

- Thread starter chemic_23
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- #1

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what is the limit of

- #2

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I would do it this way:

remember that:

a[tex]^{2}[/tex]-b[tex]^{2}[/tex]=(a+b)(a-b)

so

a-b=[tex]\frac{(a^2-b^2}{(a+b)}[/tex]

so take the numerator and denominator of your problem and treat each as a-b

so:

[tex]\sqrt{x}-3[/tex] = [tex]\frac{x-9}{\sqrt{x}+3}[/tex] (1)

and

[tex]\sqrt{1+\sqrt{x}}-2[/tex] = [tex]\frac{\sqrt{x} - 3}{\sqrt{1+\sqrt{x}}+2}[/tex] (2)

so your limit is now (1)/(2): which gives:

[tex]\frac{x-9}{\sqrt{x}+3}\times\frac{\sqrt{1+\sqrt{x}}+2}{\sqrt{x}-3}[/tex]

Is this clear?

Do you know how to continue from here?

- #3

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continuation...

thus, the limit of

- #4

Dick

Science Advisor

Homework Helper

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Correct.

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